委托构造函数和继承构造函数

委托构造函数

先来看下，我们类的构造函数不使用委托构造的情况

class AA{
    public:
        int first;
        int second;
        int third;
    
    AA() {}
    AA(int first) {
        this->first = first;
    }

    AA(int first, int second) {
        this->first = first;
        this->second = second;
    }

    AA(int first, int second, int third) {
        this->first = first;
        this->second = second;
        this->third = third;
    }
};    

可以看到上述的所有构造函数，存在重复代码

使用委托构造函数

class AA{
    public:
        int first;
        int second;
        int third;
    
    AA() {}
    AA(int first) {
        this->first = first;
    }

    AA(int first, int second) :AA(first){
        this->second = second;
    }

    AA(int first, int second, int third): AA(first, second) {
        this->third = third;
    }
};  

继承构造函数

class AA{
    public:
        int first;
        double second;
        string third;
    AA(int first, double second, string third) {
        this->first= first;
        this->second = second;
        this->third = third;
    }
};    

class Child :public AA {
    public:
        using AA::AA; // 使用using AA::AA;后，子类就可以直接使用父类的构造函数了
};

int main() {
    Child child(22, 3.14, "hello world");
    return 0;
}
``